If $\frac{60}{2^3\cdot5^8}$ is expressed as a decimal, how many non-zero digits are to the right of the decimal point?
Solution: Rewriting and reducing the fraction, we get $\frac{60}{2^3\cdot5^8} = \frac{2^2\cdot3\cdot5}{2^3\cdot5^8} = \frac{3}{2\cdot5^7}$. Multiplying the numerator and denominator by $2^6$, we obtain
\[\frac{3}{2\cdot5^7}\cdot\frac{2^6}{2^6} = \frac{3\cdot2^6}{2^7 \cdot 5^7} = \frac{192}{10^7} = .0000192.\]Therefore, there are $\boxed{3}$ non-zero digits to the right of the decimal point.